To prove that $xz = \frac{1}{2} xy$ given that $xz = yz$, let's go through the following steps:

1. Represent Points on the Number Line

Let's assume $x$, $y$, and $z$ are real numbers representing points on a number line.

2. Given Condition

The given condition is that $xz = yz$.

This implies that the distance from $x$ to $z$ is equal to the distance from $z$ to $y$.

3. Expressing the Distances

The distance between two points $a$ and $b$ on a number line is given by $|a - b|$.

So, the distances can be expressed as: $xz = |x - z|$ $yz = |y - z|$

Given that $xz = yz$, we have: $|x - z| = |y - z|$

4. Solving the Equation

The equation $|x - z| = |y - z|$ implies two possible cases:

1. $x - z = y - z$
2. $x - z = -(y - z)$

Case 1: $x - z = y - z$

If $x - z = y - z$, subtract $z$ from both sides: $x = y$ This case is trivial and does not provide any meaningful result for distinct points.

Case 2: $x - z = -(y - z)$

If $x - z = -(y - z)$, simplify: $x - z = -y + z$

Add $y$ to both sides: $x + y = 2z$

Now, solve for $z$: $z = \frac{x + y}{2}$

5. Express $xz$ and $xy$

Now that we have $z = \frac{x + y}{2}$, we can express $xz$ and $xy$ as follows: $xz = |x - z| = \left|x - \frac{x + y}{2}\right| = \left|\frac{2x - x - y}{2}\right| = \left|\frac{x - y}{2}\right| = \frac{|x - y|}{2}$ $xy = |x - y|$

6. Conclusion

Since $xz = \frac{|x - y|}{2}$ and $xy = |x - y|$, we have: $xz = \frac{1}{2} xy$

Thus, the statement $xz = \frac{1}{2} xy$ is proven.

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Would you like further details on any step, or do you have any questions?

Here are some related questions:

1. How does the concept of the midpoint relate to the result $z = \frac{x + y}{2}$?
2. What are other conditions that can be derived if $xz = k \times xy$ where $k$ is a constant?
3. How can this proof be extended to points in a plane rather than just on a number line?
4. What happens if $x$ and $y$ are complex numbers instead of real numbers?
5. How would this situation be represented on a coordinate plane?

Tip: Always consider the geometric interpretation of algebraic results, as it often provides deeper insight into the problem.